1.

What is the source temperature of the Carnot engine required to get 70% efficiency? Given sink temperature = 27^(@)C

Answer»

`1000^(@)C`
`90^(@)C`
`270^(@)C`
`727^(@)C`

SOLUTION :`eta = 1 - (T_(2))/(T_(1))`
`0.7 = 1 - (273 + 27)/(T_(1)) = 1 - (300)/(T_(1))`
`(300)/(T_(1)) = 0.3`
`T_(1) = 1000 K = 727^(@)C`


Discussion

No Comment Found

Related InterviewSolutions